Find the sum of first n squares, difference equation approach, (ft. Max!)
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- čas přidán 9. 09. 2017
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Find the sum of first n^2, ft. Max!
find 1^2+2^2+3^2+...+n^2,
difference equation,
1^2+2^2+3^2+...+n^2 formula,
1+2+3+...+n formula,
find the sum of first n squares,
find the sum of first squares of n,
sum of the first n natural numbers,
series of n^2,
blackpenredpen,
math for fun,
Loool dr payem in the end just sitting in the chair like a student
Student of the game
Title lies possible viewers solution presented here has nothing to do with difference equation approach
He used perturbation the sum, not difference equation approach
Agar bich mai multiply sign ho to or sabbi no. square ho to kya kre
The most beautiful thing is (1+2+3+4+........n)^2=(1^3+2^3+3^3+.......n)
Yes, it is!
Yes dude the formula for Sum of squares of n natural numbers is equal to the square of the formula for finding the sum of n natural numbers
@@ashagodha3630 no sum of cubes
what is this correct?
Is this a coincidence?
The only thing I'm wondering, though, is how would someone come up with using that formula? How would one figure that out? I feel like it would potentially be difficult.
he can make a conjecture and prove it by recurrence
@@mathieumourey8594 Maybe, but what would you base the conjecture on?
I believe they used the binomial expansion theorem to find out that telescoping sum. Whats amazing about using this method is that...you can pretty much know the sums 1^k + 2^k + ... + n^k, for some k is an integer! Check Mathologer's masterclass on the topic.
@@MG-hi9sh The masterclass is a bit long, about an hour. But it'll be worth it!
@@burningMallowz Hey, thanks mate! I’ll give that a try some time. Unfortunately, I have an exam tomorrow, so I can’t look at it today. 😔
I do find these expressions for summations up to the nth term very interesting.
I remember doing this. From S1 to S5, I used Pascal Triangle.
But at S6, Pascal Triangle alone can't create the equation, or at least it will take a very long time. So I manage to find a pattern for Sn, with Pascal Triangle, I solve for S6 and S7.
When I was doing S8, I discover that I can use many (n²+n) to make things easier. With Python, S8 and S9 are done.
Weeks later, I continue for S10 and S11, but this time with really big simultaneous equations (slowly but surely).
I stopped at S11, because I don't see anymore pattern of the equations for Sn.
Pattern found:
• S1 is factorisable on all Sn, n≥1,
S2 is factorisable on all Sn, n is even, S3 is factorisable on all Sn, n is odd.
• Factorising 2n+1 on all Sn, n is even, the leftovers can be expressed with n²+n, without leaving any n alone.
• After expanding everything, the term with the largest power of n for Sa is (n^(a+1))/(a+1).
I had to memorize this rule but I didn't know how it works. Today I finally understand that. Thank you.
I know, I'm so glad I found this. That proof was stunning.
me too man, the book I had just threw the conclusion to me
Wow, this derivation would have saved me a lot of trouble had I known about it. Thanks for showing this!
GOD finally a good video on the subject, thanks a lot! you explained it very clearly
Amazinnnnng, Have been looking for a demonstration of this for weeks now, while wolframathalpha will just spit the solution.
Oh my.....
This is what i have been looking for so longgggg. Thank you BlackpenRedpen
There is a tiling proof for n(n+1)(2n + 1)/6! If you can stack squares, with the smaller ones towards the corner. This can be done in Blender, Minecraft, or Tinkercad. You can also use Lego or Snap Cubes to do this. It is possible to tile 6 identical copies of these stair shapes into an (n) by (n + 1) by (2n + 1) block.
Hey, if you don't mind, we'd appreciate a link!
I love these series videos!! They're awesome
I love your channel, since i just found it couple days ago. say hi to you. I'm just an admirer of math, and a math and physic private teacher for senior hi-school. Surely it's my pleasure to watch your channel here. Somehow i feel like being on a joyfull travelling when see or solve math cases. I am lovin' it. Thank God i've found this channel. And i like the way you give explanation, the way you speak. Keep it going Bro BlackpenRedpen. ☺☺☺☺👍👍👍👍👍👍👍👍👍
This guy is really cool I'd like to see more videos with him
How did max know to use (n + 1)^3 - n^3? The whole derivation seems quite dependent on that.
it is one way to solve it, you often look for telescopics on sums so you can get rid of alternating terms and get something that you want. in this case the n^2 terms, you can apply the same concept to find a formula for the n^3 sum
@@bossdelta501but it still comes out of.nowhere and there are no alternating terms in the sum 1^2 + 2^2...+ n^2 so there is NO REASON I can see to think of cubes at all, wouldn't you agree?
I'll admit, I've never seen any of Dr. Payem's videos, but I love him every time he shows up in bprp's videos! XD
Thanks for giving such type of explanation...it helps me so much.
Yei!! .. One of my favorite subjects, hope see it soon in my career and more vídeos of its. LoveIt!
same here!!! I have a few more coming.
(Yeeei!!)x2 I wanna it soon. Greats from Colombia :)
Yeah, I love series, it's amazing math concept that shows really important relations:)
Good video. Appreciate your efforts and willingness to share knowledge
thanks for explaining! this is great!
I watched this last year but, I am revisiting because it came up again! thanks.
Great work 💙✌
Keep it up
niiice
I was looking for a nice evaluation of this sum
Keep it going bro you are too good
Really liked this video. Love the guest speakers!
arraysstartat1 thank you!!! I will have more in the future!
In case anybody is wondering, the general formula is kind of a nuance, but it can be compactly written in a way such that a computer would be able to compute it. For example, if S_k(n) is the sum of the first n integers raised to the kth power, then:
S_k(n) = {n^(k+1) - sumj_(1, n) [(-1)^(k-j) * (k+1 chose j) * S_j(n)]}/(k+1)
^^^That would pretty much be the majority of the code necessary for, say, Haskell to compute your value. If you're using something stronger, like Mathematica, then I'm sure this forumla is already in a library somewhere.
There is an alternative way of doing this that takes less time for x^k, you can show by bring the differences of these functions (x + 1)^k - x^k and taking the differences of that until it's a constant, taking every term till it's a constant and it will mimic how long it takes for a polynomial to go to a constant this same way. Theres more too this that is hard to fit into a comment but you can take k + 1 terms to the summation of x^k from 1 to n in regression (plugging the x and y values in for each and solving the variable system), this generates the coefficents and makes the polynomial function.
Amazing
Thank you for sharing the logic behind this video
This was given in Class 11th NCERT
You explained it awesome
very well explained.
Great video. Thank you very much.
Very different point of view thanks
Simply stunning :) it is beautiful and elegant the way he derived the summation formulas 😶
Yea!!! Dr. Peyam and I liked it very much too.
@@blackpenredpen , I am a math grad myself, but what amazes / mystifies me about the proofs / derivations of mathematical many results is, how in the world do you START ?? And having started, how do you PROCEED !! I mean - just how do you know what expressions you have to manipulate and how do you know how to manipulate ?? I learnt to "prove" these two results by using principle of mathematical induction. And these algebraic proofs [esp., the second one ] are totally amazing !! For example, how did you know that if you re-wrote S as n +(n-1) +(n-2) +...2 +1, it would help you later ?? Okay, forget this - this atleast is only one step away from what you want to find out. In the second part,, how did you know you had to manipulate (m+1)^3 - m^3 ?? The expression you are trying to evaluate is too far from where you started !! NONE OF MY PROFESSORS / LECTURERS HAVE BEEN TO ABLE TO QUELL MY AMAZEMENT - in fact, I get snubbed [by even fellow students] for "wasting time" !!
@@blackpenredpen yea but whonwould ever think of doing this?? I hope you can please please respond and tell me.this is not intuitive or logical at all..
@@muralinagarajan8305 exactly this is not intelligent or logical or intuitive no NO ONE WOULD EVER THINK TO DO THIS to prove it..so gow would someone really. If no one showed you how..not even Ramanujan or me or somebody would think of that
@@muralinagarajan8305 Oh, that's simple. These things happen because, has you go through your daily routines, you might get lost inside your mind. You might think that you're only ever doing math when, you know, you have a piece of paper and is actly doing math. That couldn’t be fader from the truth. If one has a really playful mind and a sense of creativity, even as you step back from a problem, if you weren’t able to solve it immediately, your mind might be still processing it on the background. If you go back to the problem later, you may carry with you a new, fresh perspective. If you are really creative and really knows how to exploit this system to it's logical extreme, people can come up with really intelligent unexpected solutions. So these incredible proofs are, really, at their core, a feat of creativity by persons way more ingenious than you and me. Let me tell you, a big reason why people can get such staggering results is that their subconscious mind do half the job for then.
In other word, what I'm saying is, beyond being incredible logical feats on their on, they're also displays of amazing creativity. Be warned: math can be also a form art sometimes.
Guys I think I've found the pattern!
S(0) = n
S(1) = [n(n+1)]/2
S(2) = [n(n+1)(2n+1)]/6
S(k) = {n[n+1][2n+1]...[(k-1)n+1][kn+1]}/(n+1)!
So I think the sum of all n cubes will be (n)(n+1)(2n+1)(3n+1)/24
love your videos, bprp :3
(P.S. at the time of this comment, I'm 12)
@Zavion W. / BlockyKirby314
unfortunately, this does not hold.
sum of cubes from 1 to 2 = 1^3 + 2^3 = 1 + 8 = 9.
if we plug in n=2 into your formula, we get 2(3)(5)(7)/24 = 35/4.
Close, but no banana.
Keep at it, though. At 12, you've got a world of mathematics magic ahead of you. Enjoy the ride!
@@jcb3393 Thank you very much! It's good to know that at least I was close lol 😅
I turned 13 in December btw 😃👍
@@zavionw.8052 So what? What do you mean by adding how old you are in here? Is it relating to the question? Or you just wanna show up sth? Be humble boy, there are millions of geniuses out there who younger than you are. Stay hungry stay foolish!
@@phucl.nguyen5168 dam relax dude. hes just a kid that feels proud. he doesnt always have to feel like trash bcuz others are better than him. wonder what ur childhood was like...
@@OriginalEch3Official Ofc thats true. I mean, he can do good when he's 12. He can't like change the past honestly. He has to accept his past and move on and try to do better. Who knows? maybe he'll end up doing something great!
I have always been trying to derive power formulas by myself to challenge myself, but I couldn't find a nice and neat way to derive the sum of squares. This is a very interesting method, very nice :)
Khan Academy have clear explanations of this sum. :)
It's ez to prove tbh
@@peamutbubberbut how. It's not at all ez to derive..proving is not a simportatn..indont see why anyone would thinknof difference of twomcubes?
Dude, awesome job :)
thanks:)
very very cool. always wondered how that formula was derived and why it works. this perfectly satisfies my curiosity. thank you!!
agreed. just the comment I was looking for.
Thank you
It helped a lot
I think it’s very interesting that S4 (sum of finite cubes) is the formula for S2 just squared.
I found it to be -1/30(n) + 1/3(n^3) + 1/2(n^4) + 1/5(n^5) using matrices and reduced echelon form.
Good.....mantaaap...hebaattt
Dude.. Awesome 👌
aah old memories of arithmetric progression and series came in mind. Learned this when i was in class 9. I remember finding out the sum of cubes of n number by same method. That was amazing. I thought I discovered something. Hahaha.
it helped me a lot
Excellent video !
Super.....explanation nice simple....good.....great sir
Really excellent...
Great one!
So nice derivation to find sum of square natural no series... Thanks
Yup!
Below you will find the Mathematica line to calculate the coefficients for the first 10 polinomials:
Inverse[Table[If[m>n-1,0,Binomial[n,m] (-1)^(n+m+1)],{n,1,k},{m,0,k-1}]/.k->10]//MatrixForm
Beautiful!!!
thank you very much from Algeria.
Great! Thanks guys!
This is great
Awesome!
I actually found another elegant way you can find the second sum in the video a little bit more complicated if C is our second sum of n numbers squared basically if you take the differemce of a number squared minus it s former squared except for 1 you actually see this for example 2squared -1squared=3 also 3squared-2squared=3+2 4squared-3squared=3+4 5squared-4squared=3+6 now if we set a=3 for convension purposes and solve for its number we will see this 2squared=a+1 3squared=2a+3 4squared=3a+7 so if C=1+2squared+3squared...nsquared=1+a+1+2a+3+3a+7+4a+13... now if A=the first sum of the video then C=1+a+1+2a+3+3a+7+4a+13...=1+a(A-n)+1+3+7+13...=1+a(A-n)+1+(1+2)+(1+2+4)+(1+2+4+6)... now we can clealry see a pattern 1+a(A-n)+1×(n-1)+2×(n-2)+4×(n-3)+6×(n-4)...=n+a(A-n)+2×(n-2+2×(n-3)+3×(n-4)+4×(n-5)...) now is the very tricky part that need you need to pay very much attention or you will get lost we know that n-2+n-3+n-4+n-5...=A-n-(n-1)lets name this B then n+a(A-n)+2×(n-2+2×(n-3)+3×(n-4)+4×(n-5)...)=n+a(A-n)+2(B+B-(n-2)+B-(n-2)-(n-3)+B-(n-2)-(n-3)-(n-4)...)=n+a(A-n)+2((n-2)×B-(n-2)×(n-3)-(n-3)×(n-4)-(n-4)×(n-5)...)=n+a(A-n)+2((n-2)×B-(n-2)×(n-2-1)-(n-3)×(n-3-1)...) now if we get rid of the brankets something magical happens n+a(A-n)+2((n-2)×B-(n-2)×(n-2-1)-(n-3)×(n-3-1)...)=n+a(A-n)+2×((n-2)B-(n-2)squared+n-2-(n-3)squared+n-3-(n-4)squared+n-4...) but we know that C=nsquared+(n-1)squared+(n-2)squared... and B=n-2+n-3+n-4+n-5... this means that n+a(A-n)+2×((n-2)B-(n-2)squared+n-2-(n-3)squared+n-3-(n-4)squared+n-4...)=n+3A-3n+2((n-2)B+B-C+(n-1)squared+nsquared)= n+3A-3n+2((n-2)B+B-C+(n-1)squared+nsquared)=-2n+3A+2×(n-1)B-2C+2(n-1)squared+2nsquared now lets replace our varyables -2n+3A+2(n-1)(A-(n-1)-n)-2C+2(n-1)squared+2nsquared=-2n+3A-2(n-1)squared-2n(n-1)+2(n-1)A+2nsquared+2(n-1)squared-2C=-2n+3A+2nsquared-2nsquared+2n+2(n-1)A-2C=3A-2A+2nA=A×(2n+1)-2C but since this is C then 3C=A×(2n+1) C=A×(2n+1):3=n×(n+1)÷2×(2n+1)÷3=n×(n+1)×(2n+1)÷6 and there you go trust me it s much more easyer on paper also let me know if this proof actually exists cause I figured it out myself
Interpolation, recurrence relation with generating functions, using differences (discrete version of calculus),
I prefer the khan academy method about patterns of patterns and ended up like 2/6 n'3 + 3/6 n'2 + 1/6 n , which is the known formula of: n(2n+1)(n+1)/6. I like it because youdont need to know anything other formula like this video.
天啊這個 Maks 的英文我真的聽得懂耶
黑紅筆抱歉了 我其實常聽不清楚你的英文, 雖然你教得真的很好
There are more methods to prove this, when I proved this to myself I used the fact that (1+3)=2^2 (1+3+5)= 3^2, generalized it's the partial sum of the odd numbers up to the i-th number of i^2 (odd1+odd2+odd3+....+oddi)=i^2, the rest is trivial because you have n + 2(n-1) + .... + n(n-(n-1)) + n(n-n).
love this
This has all been formalized in the calculus of finite differences, a subject that is powerful and beautiful but is little known.
or in this topic = book = Concrete Mathematics_ A Foundation for Computer Science
Very cool! Is there a general formula for any k? Or even further, an extended formula for any k being a real number?? @blackpen
I've just realized that this is somehow related to the Riemann zeta function... It has 'almost the same form' if n goes to infinity (for the part that is not an analytic continuation).
Idk if it holds for every k but I think I spot a pattern.
For k=0, n(0n+1)/1
For k=1 n(0n+1)(1n+1)/1*2
For k=2 n(0n+1)(1n+1)(2n+1)/1*2*3
If I am right every time k increases by 1 you multiply by (kn+1)/(k+1).
I honestly don't know about real k
Recently, I have discovered a way to derive this formula geometrically. However, the math to get there is a bit more tedious, but it works. I would be very willing to share it if you are curious. (On a side note: I also found a geometric way to derive the formula for the sum of the first n perfect cubes.)
Great to see more Penny!!
Penny? U mean Peyam?
blackpenredpen No. Not him. You! I coined Penny from your username BlackPenRedPen
OH! I see, lol.
So, instead of Pen, it's Penny?
blackpenredpen indeed!
Ah, I see!
that was elegant
Another way is just assume that S2(n)=a*n^3+b*n^2+c*n+d. Calculate it for n=1,2,3,4; we will have 4 linear equations with 4 variables. Solve it, a=1/3 b=1/2 c=1/6 d=0. We will have same formula. Exactly same way to calculate S_k(n).
Nice catch! :)
Not as elegant, but doesn't require leaps of insight. From the formula of arithmetic progression one can suspect that it's a polynomial of degree higher by one than the degree of the series.
It's very cool!
Bri Lianna thank you!!
At 3:38, why do we use (n+1)^3-n^3? How do we arrive at this identity? Thanks!
(May be a bit late) We don't arrive at it. We just take a known formula that we can use to our advantage. In this case we take (x+1)^3 = x^3−3x^2+3x−1 and then we remove the x^3 because we don't need it.
@@grivar
Except that (x+1)^3 = x^3 *+* 3x^2 + 3x *+* 1.
Your right-hand side is equal to (x-1)^3.
So would you basically just guess that the result of S_n will involve x^(n+1), and then consider how you get from any natural number x to its successor?
(After all, all of this is proven by induction.)
Yeah! I’m lost too, seems like a convenient step with a lot of trial and error behind it, I think it would be super cool to see the blind alleys and failed ideas that led to an uglier version of these identities only to be simplified, I want to see the process behind it, nor just this rehearsed explanation of definitive steps
Nice video 👌👌👍👍
I loved it.... great... love from india..😍😍
Thank you
CHALLENGE: Using a similar method, it is not so hard to prove that S3 = [m^2(m+1)^2]/4 = (S1)^2. I can easily derive this by algebra BUT can anyone come up with a GEOMETRIC proof for this? It SEEMS like there MUST be a geometric proof for this, but I have never seen it.
OH OH!! I KNOW I KNOW!!
Since I wont be able to make videos till later. Here's a hint for you.
Count the number of rectangles in a , let's say, 8x8 chessboard. : )
The number of SQUARES on an n x n chessboard is indeed n(n + 1)(2n + 1)/6 = S2, but this is not what I asked. I asked if there was a geometric proof for S3 = (S1)^2 which has nothing to do with S2.
take three cases, 1^3, 1^3+2^3, and 1^3+2^3+3^3 then for each case draw a square, and you will see something interesting:)
Yes, Yes. I have now created a geometric proof. Imagine that you have a square that has side length of n(n+1)/2 filled with unit cubes and we add a cube to it constructed of (n + 1)^3 unit cubes. This will make the total number of cubes [n(n+1)/2]^2 + (n + 1)^3. Expanding and simplifying we get (n^4 + 6n^3 + 13n^2 + 12 n + 4)/4 which factors to [(n+1)(n+2)/2]^2 which is the SAME expression you would get for the number of cubes in a square that is (n+1)(n+2)/2 on a side which is exactly n + 1 more on a side than the square we started with.
Let's use some integers to make this easier to visualize. Let's make n = 4. The sum of integers from 1 to 4 is n(n+1)/2 = 4(4+1)/2 = 10. Construct a square containing [n(n+1)/2]^2 = 10^2 = 100 cubes (10 on a side). Now we wish to add to this, a cube that is n + 1 = 5 unit cubes on a side so it has a total of (n + 1)^3 = 5^3 = 125 unit cubes. We can now show that from all of these cubes we can construct a square that is 15 cubes (= the sum of integers from 1 to 5) on a side or 15^2 = 225 cubes and this is because [n(n+1)/2]^2 + (n + 1)^3 = [(n+1)(n+2)/2]^2.
You can use induction to come up with a proof too. When n=1, S3=1^2=(S1)^2. Let's say you have a square of side length s (to mean S1 for tidyness sake), and you want to create a square with side length s+n+1, you would need to add 2s(n+1)+(n+1)^2 units to the area. Substituting s=n(n+1)/2, you are adding (n(n+1)(n+1)+(n+1)^2) to the area, or simplifying it, (n+1)^3 to the area (originally s^2). Therefore s^2+(n+1)^3=(s+n+1)^2.
Wow! The sum of cubes was pretty hardcore!
Awesome
Sublime!
I bet coming up with math techniques must be the hardest thing in the world.
Pascal's identity, nice
I love this video, the sums are simple, but he provides us with elegant solutions with a similar enthusiasm to you 😃
Thanks:)
Thanks!
Yes but what's the general form for any exponent?
it's coming soon:)
thank you bro
What if k=n?
So 1^n+2^n+3^n+…+n^n = ?
www.wolframalpha.com/input/?i=Sum+from+1+to+n+of+k%5En
See here: en.wikipedia.org/wiki/Faulhaber%27s_formula#Examples
MamboBean
(n(n+1)(2n+1)(3n+1)(4n+1)... (n^2+1))/n!
I think
Could also maybe be:
(n(1!n+1)(2!n+1)(3!n+1)... (n!n+1))/n!
It's called Faulhaber Formula, look for it!
You use the same method. If you watch his video on deriving the sum of cubes, it'll all make sense.
czcams.com/video/fw1kRz83Fj0/video.html
妙啊(marvelous!) >.
I'm not sure I understand this method, but I think it is similar to what we learnt for sigma notation. Nevertheless, great video.
I found a recursive equation for finding the anti difference of n^m for any positive integer m :) I just need to make one step in the proof a little more rigorous
I always tried using the cubic pyramid method, for sN, find the volume of a stepped pyramid with a base of side length m and N dimensions. So for s3, it is a pyramid with a cube base (4d shape) but this new method is much better looking
thanks
nice one.
Dude, you are a wizard. Do you have any background on how the person (or even you) who came up with this derivation rationalised the selection of such a process? Like, how did they think of starting with the cube of n?
The difference (n+1)^k - n^k is always a polynomial of degree k-1 because it has the term k•n^{k-1} and the terms n^k eliminate each other.
If you note that, you can represent the polynomial a_n = n^k as the difference a_n = b_{n+1} - b_n, where b_n is certain polynomial of degree k+1.
b_n could be found as an arbitrary polynomial of degree k+1 (depends on k+2 coefficients), whose difference b_{n+1} - b_n is exactly n^k.
Then, the sum of a_n is b_{n+1} - b_1 (as shown in this video).
If a_n = n^2, then b_n could be chosen as
n^3/3 - n^2/2 + n/6 + c,
where c is an arbitrary number.
For c=0 it can be rewritten as
(n-1)n(2n-1)/6.
Then the sum is b_{n+1} - b_1 = n(n+1)(2n+1)/6.
This is the logic of the method. But it could be difficult to see, since the general method was adapted and applied to a specific example.
Sk can be expressed as a polinomial of degree k+1.
The general case depends on Bernoulli numbers, but each particular case can be solved with a linear system of k+1 equations.
S2 is particularly easy because you can use some rules:
* the coefficient of n^(k+1) is always 1/(k+1).
* the coefficient of n^k is always 1/2.
* the constant term is always 0.
* all the coefficients add up to 1.
So you immediately get that
Sum[t^2,{t,1,n}]=
1/3 n^3 +1/2 n^2 +1/6 n
Other useful rules:
* all polinomials have roots in 0 and -1.
* all coefficients of powers k-2m (m>0) are 0.
The explanation is awesome, but how does it come to mind to take cube of( n-1) to substract cube of n... That's the most important thing to understand, please make us understood if you can
I noticed that when squared, the Integer Sum Formula results in the k=3 case. Is this a coincidence?
It's special, since no other two Sn share that type of relation.
@@assiddiq7360 :
Is it a coincidence though?
thanks sir
I wonder if there’s a derivation for a general formula for S_k
That looked like easy money then the S2 showed up
awesome
gee thanks bro
The sum for S₂ comes up in summing over the Zeeman energy levels in working out Curie's Law for paramagnets in statistical thermodynamics of magnetism. I came up with a straightforward - and I think more systematic - way of doing the sum as a simple linear algebra problem: Assume that f(N) = 1² + 2² + ... + (N-1)² + N² is a cubic expression since there are N terms and the general form of each term is quadratic. Then f(N) = a₀ + a₁N + a₂N³ + a₃N³. Since f(0) = 0, a₀ = 0. You can find equations involving the coefficients a₁, a₂, and a₃ by plugging in N = 1, 2, and 3 to get f(1), f(2), and f(3). Now write a matrix equation 𝑷𝐚 = 𝐟 where 𝑷 comes from the equations, 𝐚 is the column vector (a₁, a₂, a₃) and 𝐟 is the column vector (f(1), f(2), f(3)). Invert 𝑷, and multiply 𝑷⁻¹ by both sides of the matrix equation to get the coefficients, which you use to get f(N). This is generalizable to higher power sums, of course.
Now im wondering if this method can be extended into higher powers...
Like finding S3 using (n+1)^4 - n^4
Idk if it holds for every k but I think I spot a pattern.
For k=0, n(0n+1)/1
For k=1 n(0n+1)(1n+1)/1*2
For k=2 n(0n+1)(1n+1)(2n+1)/1*2*3
If I am right every time k increases by 1 you multiply by (kn+1)/(k+1)
@@justanormalyoutubeuser3868 oh wow! Never noticed that before! Nice catch!
Is there a general formula for any sum of powers? I found the next three expressions, but there does not appear to be a pattern:
S0 = n
S1 = n(n+1)/2
S2 = n(n+1)(2n+1)/6
S3 = n^2(n+1)^2/4
S4 = n(n+1)(2n+1)(3n^2+3n-1)/30
S5 = n^2(n+1)^2(2n^2+2n-1)/12
The only thing I have noticed is that when you expand each expression then the two highest powers in n always follow
Sk = n^(k+1)/(k+1) + n^(k)/2 + O(n^(k-1))
K van der Veen, there is a general expression, but is not nice and involves Bernoulli's numbers.
K van der Veen, somebody posted a link to Wikipedia for the article "Faulhaber formula"... very interesting reading!
S3=(S1)^2
my maths teacher dreived the formulas in the exact same way lol
Thank you so mush in moroco